JYU Applied Geochemistry & Lab Ch.7 Redox Reactions Part 1.

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Presentation transcript:

JYU Applied Geochemistry & Lab Ch.7 Redox Reactions Part 1

 Oxidation state: the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic  Oxidation: Increase in oxidation state (of an atom, molecule, or ion)  Reduction: Decrease in oxidation state  Redox reactions: Reactions involving oxidation and reduction among the reactants due to electron transfer 1. Definitions

2. Theories

 Significance  Prediction of the stable redox species  Procedures  Calculation of water stability limits  Setting up the system  Construction redox reactions among the possible species  Application of the Nernst equation  obtaining the relation between pH and E H for each reaction  Plot the pH-E H relation and define the stability field for each of the species on the diagram 3. pH-E H Diagram

 Water stability limits  Upper limit  0.5O 2 + 2H + + 2e - = H 2 O  E=E o /2 log [H+]2P o2 0.5  E= pH (P o2 =1 atm)  Lower limit  2H + + 2e - = H2  E=E o /2 log [H+]2/P H2  E=-0.059pH (P H2 =1 atm)

Outlined area shows the region delineated by Bass Becking et al. (1960) for a large number of measurements of natural waters. Small circles are ground-water samples (from Fish, 1993).

 Fe-O-H system  Possible species  Fe 2+, Fe 3+, FeOH +, FeOH 2+, Fe(OH) 2 +, Fe(OH) 3 0, Fe(OH) 4 -, (FeOOH(am), Fe(OH) 3 )  Data Species  G o f at 298K & 1bar Fe Fe FeOH Fe(OH) Fe(OH) Fe(OH) FeOH H2OH2O

 Reactions  Fe 3+ - Fe 2+  Fe 3+ + e - = Fe 2+ b   G o f = = (kcal/mole)   E o = 17.73/23.06=0.769 (v)  E= log([Fe 2+ ]/[Fe 3+ ])=0.769  FeOH 2+ - Fe 2+  FeOH 2+ + H + + e - = Fe 2+ + H 2 O   G o f = = (kcal/mole)   E o = /23.06=0.898 (V)  E= pH

 Fe(OH) Fe 2+  Fe(OH) H + + e - = Fe H 2 O   G o f = * = (kcal/mole)   E o = /23.06=1.120 (V)  E= pH  Fe(OH) Fe 2+  Fe(OH) H + + e - = Fe H 2 O   G o f = * = (kcal/mole)   E o = /23.06=1.401 (V)  E= pH

 Fe(OH) Fe 2+  Fe(OH) H + + e - = Fe H 2 O   G o f = * = (kcal/mole)   E o = /23.06=1.920 (V)  E= pH  Fe 3+ - FeOH 2+  Fe 3+ + H 2 O = FeOH 2+ + H +   G o f = =2.977 (kcal/mole) = * * log[H + ]=1.364pH  pH=2.183

 FeOH 2+ - Fe(OH) 2 +  FeOH 2+ + H 2 O = Fe(OH) H +   G o f = =5.117 (kcal/mole) = * * log[H + ]=1.364pH  pH=3.751  Fe(OH) Fe(OH) 3 0  Fe(OH) H 2 O = Fe(OH) H +   G o f = =6.487 (kcal/mole) = * * log[H + ]=1.364pH  pH=4.756

 Fe(OH) Fe(OH) 4 -  Fe(OH) H 2 O = Fe(OH) H +   G o f = = (kcal/mole) = * * log[H + ]=1.364pH  pH=8.773  Fe 2+ - FeOH +  Fe 2+ + H 2 O = FeOH + + H +   G o f = = (kcal/mole) = * * log[H + ]=1.364pH  pH=8.999

 Fe(OH) FeOH +  Fe(OH) H + + e - = FeOH + + 3H 2 O   G o f = * = (kcal/mole)   E o = /23.06=1.388 (V)  E= pH